Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}
using ll = long long;

struct segtree{
	using D = int;
	D inf = 1e9;

	int N;
	vector<D> val;

	segtree(){}
	segtree(int n){
		N=1;
		while(N<n) N*=2;
		val.assign(N*2,inf);
	}
	D getval(int k){
		k+=N;
		D res = inf;
		while(k){
			chmin(res,val[k]);
			k/=2;
		}
		return res;
	}
	void changemin(int a,int b,D d){
		a+=N,b+=N;
		while(a<b){
			if(a&1){
				chmin(val[a],d);
				a++;
			}
			if(b&1){
				b--;
				chmin(val[b],d);
			}
			a/=2,b/=2;
		}
	}
};


int N;
ll K;
const int MN = 100010;
ll inf = 1e18;
ll A[MN],F[MN];
ll B[MN];
using P = pair<ll,ll>;

ll solve(vector<P> vp){
	int N = vp.size();
	if(N==0) return 0;
	vector<ll> xs;
	xs.pb(0);
	xs.pb(K);
	rep(i,N){
		xs.pb(vp[i].fs);
		xs.pb(vp[i].sc);
	}
	sort(all(xs));
	xs.erase(unique(xs.begin(),xs.end()),xs.end());
	auto getid = [&](ll x){
		return lower_bound(all(xs),x) - xs.begin();
	};
	int X = xs.size();
	segtree seg(X);

	vector<ll> dpl(N,inf),dpr(N,inf);
	for(int i=N-1;i>=0;i--){
		int l = getid(vp[i].fs), r = getid(vp[i].sc);
		{
			int j = seg.getval(l);
			if(j>1e8){
				dpl[i] = 0;
			}else{
				ll dis = vp[j].fs - vp[i].fs;
				if(dis<0) dis += K;
				dpl[i] = dpl[j] + dis;
			}
		}
		{
			int j = seg.getval(r);
			if(j>1e8){
				dpr[i] = 0;
			}else{
				ll dis = vp[j].fs - vp[i].sc;
				if(dis<0) dis += K;
				dpr[i] = dpl[j] + dis;
			}
		}
		if(l<=r){
			seg.changemin(0,l,i);
			seg.changemin(r+1,X,i);
		}else{
			seg.changemin(r+1,l,i);
		}
	}
	ll res = inf;
	rep(i,N){
		int l = getid(vp[i].fs), r = getid(vp[i].sc);
		{
			int j = seg.getval(l);
			if(j>i) chmin(res,dpl[i]);
		}
		{
			int j = seg.getval(r);
			if(j>i) chmin(res,dpr[i]);
		}
	}
	return res;
}

int main(){
	cin>>N>>K;
	rep(i,N){
		cin>>A[i];
		int b;
		cin>>b;
		F[i] = b==2;
	}
	rep(i,N){
		if(!F[i] && A[i]*2>K){
			puts("-1");
			return 0;
		}
	}
	rep(i,N){
		B[i+1] = B[i] + A[i]*2;
	}

	vector<P> vp;
	rep(i,N) if(!F[i]){
		ll l = ((-B[i])%K+K)%K;
		ll r = ((-B[i+1])%K+K)%K;
		vp.pb(P(l,r));
	}
	ll ans = B[N] + solve(vp);
	cout<<ans<<endl;
}